Question 972841
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Your first clue is to look at the coefficients of your given polynomial:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^{12}\ +\ 4x^9y^3\ +\ 6x^6y^6\ +\ 4x^3y^9\ +\ y^{12}]


Which are *[tex \LARGE 1,\,4,\,6,\,4,\,1]


And this looks exactly like either the Number 4 row of Pascal's Triangle, or the sequence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ {4 \choose 0},\ {4 \choose 1},\ {4 \choose 2},\ {4 \choose 3},\ {4 \choose 4},\ ]


one of which I will hope makes sense based on your particular understanding of the binomial theorem.  Given that, we know from the binomial theorem that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\alpha\ +\ \beta\right)^4\ =\ \alpha^4\ +\ 4\alpha^3\beta\ +\ 6\alpha^2\beta^2\ +\ 4\alpha\beta^3\ +\ \beta^4]


Let's use a substitution to turn your original mud fence into something not quite so butt ugly.


Let *[tex \Large \alpha\ =\ x^3] and *[tex \Large \beta\ =\ y^3]


Now substitute.  Since *[tex \Large x^{12}\ =\ (x^3)^4] and *[tex \Large x^{9}\ =\ (x^3)^3] and *[tex \Large x^{6}\ =\ (x^3)^2] (and similarly for the *[tex \Large y] terms), we can substitute *[tex \Large \alpha] and *[tex \Large \beta] into your original polynomial to get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \alpha^4\ +\ 4\alpha^3\beta\ +\ 6\alpha^2\beta^2\ +\ 4\alpha\beta^3\ +\ \beta^4]


which we know from earlier to be equal to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\alpha\ +\ \beta\right)^4]


Now just substitute back again to get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x^3\ +\ y^3\right)^4]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \